Problem: $\lim_{x\to-\infty}\dfrac{6x^2-x}{\sqrt{9x^4+7x^3}}=$
Answer: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x^2$, let's divide by $x^2$. In the denominator, let's divide by $\sqrt{x^4}$, since for any value, $x^2=\sqrt{x^4}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{6x^2-x}{\sqrt{9x^4+7x^3}} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{6x^2-x}{x^2}}{\dfrac{\sqrt{9x^4+7x^3}}{\sqrt{x^4}}} \gray{\text{Divide sides by }x^2=\sqrt{x^4}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}\dfrac{\dfrac{6\cancel{x^2}}{\cancel{x^2}}-\dfrac{1\cancel x}{\cancel x \cdot x}}{\sqrt{\dfrac{9\cancel{x^4}}{\cancel{x^4}}+\dfrac{7\cancel {x^3}}{\cancel {x^3}\cdot x}}} \\\\ &=\lim_{x\to-\infty}\dfrac{6-\dfrac{1}{x}}{\sqrt{9+\dfrac{7}{x}}} \\\\ &=\lim_{x\to-\infty}\dfrac{6-0}{\sqrt{9+0}} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{6}{\sqrt{9}} \\\\ &=\dfrac{6}{3} \\\\ &=2 \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{6x^2-x}{\sqrt{9x^4+7x^3}}=2$.